 Solution

For a point, stationary proton, the leading hydrogen fine structure comes from three order-$1/c^2$ contributions to the nonrelativistic electron Hamiltonian:These are the relativistic kinetic-energy correction, spin-orbit interaction and Darwin term. The spin-orbit coefficient includes the Thomas precession factor of one half; $⟨\mathbf{L}\cdot \mathbf{S}⟩={\hslash }^2[j(j+1)-l(l+1)-3/4]/2$. The Coulomb Darwin contribution is a positive contact term, relevant for S states. These corrections distinguish total angular momentum $j=l\pm 1/2$, while leaving magnetic substates degenerate without an external field.

Writing ${\alpha }_{\mathrm{fs}}=e^2/(4\pi {ϵ}_0\hslash c)$, the combined hydrogen fine-structure correction isThus the splitting is of order ${\alpha }_{\mathrm{fs}}^2$ times the principal binding energy. In the $n=2$ shell, $2S_{1/2}$ and $2P_{1/2}$ remain degenerate in this approximation, while $2P_{3/2}$ is higher by $m_e{c}^2{\alpha }_{\mathrm{fs}}^4/32$. The Lamb shift lifts the remaining S–P degeneracy; hyperfine structure is a separate nuclear-spin effect. Neither is part of the displayed correction. Nuclear recoil and finite nuclear size have also been neglected. ^[ref]